· H2 Mathematics — Ten-Year Series, worked ← All 11 questions

2019 A-Level · Paper 2 · Question 10

2019 H2 Maths Paper 2, Question 10

Correlation and regression 13 marks transformed models
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Part (i)(a)

2 marks

The five data points and Abi’s line y=35−13xy = 35 - \frac{1}{3}x are plotted below.

Part (i)(b)

1 mark

For each point, the vertical segment down to (or up to) the line is the residual for that point, also shown on the diagram.

Sketch for question 10

Part (i)(c)

1 mark

For each xx, the value predicted by Abi’s line is y^=35−13x\widehat{y} = 35 - \frac{1}{3}x, and each residual is y−y^y - \widehat{y}.

xx 40 45 50 55 60
y^\widehat{y} y−y^y - \widehat{y} 65/365/3 1/31/3 2020 00 55/355/3 −1/3- 1/3 50/350/3 1/31/3 1515 11

Sum of the squares of the residuals:

(13)2+02+(−13)2+(13)2+12\left( \frac{1}{3} \right)^{2} + 0^{2} + \left( - \frac{1}{3} \right)^{2} + \left( \frac{1}{3} \right)^{2} + 1^{2}
=43= \frac{4}{3}

∴\therefore the sum of the squares of the residuals for Abi’s line is 43\frac{4}{3}.

Part (i)(d)

1 mark

Explanation: Residuals above the line are positive and residuals below the line are negative, so in the sum of the residuals these can cancel out, giving a small (or even zero) total even when individual points lie far from the line. Squaring first removes the signs, so the sum of squares only shrinks when every point genuinely lies close to the line, making it a meaningful measure of the overall goodness of fit.

Part (ii)

1 mark

Since Bhani’s sum of squares of residuals, 11, is less than Abi’s, 43\frac{4}{3}, Bhani’s line gives a better fit to the data.

Part (iii)

1 mark

∴\therefore the least squares regression line must pass through (x‾, y‾)\left( \overline{x},\ \overline{y} \right).

Explanation: Here x‾=50\overline{x} = 50 and y‾=18.6\overline{y} = 18.6, so the line passes through (50, 18.6)(50,\ 18.6).

Part (iv)

3 marks

By GC, the equation of the least squares regression line of yy on xx is

y=−0.3x+33.6.y = - 0.3x + 33.6.

The product moment correlation coefficient is

r=−0.985 (3 s.f.).r = - 0.985\ (3\ s.f.).

Part (v)

2 marks

Using the regression line, at x=30x = 30,

y=−0.3(30)+33.6y = - 0.3(30) + 33.6
=24.6.= 24.6.

∴\therefore the estimated fuel consumption at a speed of 30 km/h30\ km/h is 24.6 km/litre24.6\ km/\text{litre}.

Explanation: This value is not expected to be reliable, since x=30x = 30 lies outside the range of the given data (40≤x≤6040 \leq x \leq 60), so the estimate requires extrapolation, and there is no evidence that the linear model still holds at that speed.

Part (vi)

1 mark

∴\therefore all of Cerie’s data points lie exactly on her regression line, i.e. the points are perfectly collinear.

Explanation: A sum of squares of residuals of 00 means every residual is 00, so every data point lies exactly on the line.

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